Lecture 10: Complexity Theory -- Space Complexity

Let M be a standard DTM. M runs in space \(S(n)\) if, for any input of length \(n\), M uses at most \(S(n)\) tape cells.

Let N be a standard NTM. N runs in space \(S(n)\) if, for any input of length \(n\), every branch of N uses at most \(S(n)\) tape cells.

PSPACE = \(\{A | A\) is decided by some DTM using a polynomial amount of space \(\}\)

NPSPACE = \(\{A | A\) is decided by some NTM using a polynomial amount of space \(\}\)

EXP = \(\{A | A\) is decided by some DTM using within 2^p(n) time \(\}\)

Because the number of configurations of a DTM is at most 2^p(n), where p(n) is a polynomial.

More specifically, the number of configurations of a DTM is at most 2^p(n) because the number of tape cells is at most p(n)

• P \(\subseteq\) PSPACE \(\subseteq\) EXP

  If a problem can be solved in polynomial time, it can be solved in polynomial space.

• NP \(\subseteq\) PSPACE \(\subseteq\) NPSPACE

NP \(\subseteq\) PSPACE 1. \(\exists\) NTM N that decides A in polynomial time

2. Simulate every branch of N : Polynomial space

3. Mark 分支的选择情况(每个结点一个)：Polynomial space()

Theorem 1

NPSPACE = PSPACE

Savitch's Theorem

Space Hirachical Theorem

\({\displaystyle {\mathsf {SPACE}}\left(o(f(n))\right)\subsetneq {\mathsf {SPACE}}(f(n))}\)

Construct a DTM D

1. D decides some language A in space \(O(f(n))\)
2. For any DTM M that decides A in space \(o(f(n))\)

D and M differs on at least one input

We want to construct D such that

    M1  M2 M3  M4  M5  M6  ...
M1  +  
M2  -   +
M3  -   -   -
M4  -   -   -   +
M5  -   +   -   -   -
M6  -   -   -   -   -   +
...
D   -   -   +   -   +   -   ...

D runs in space \(O(f(n))\) D decides some language -- will halt

Time Hierarchy Theorem

\({\displaystyle {\mathsf {DTIME}}\left(o\left(f(n)\right)\right)\subsetneq {\mathsf {DTIME}}(f(n){\log f(n)})}\)

• Have to update COUNTER every time -- \(log f(n)\) time

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最后更新: 2024年12月16日 00:12:48
创建日期: 2024年12月16日 00:12:48